Every chemical element goes through natural exponential decay, which means that over time its atoms fall apart. The speed of each element's decay is described by its half-life, which is the amount of time it takes for the number of radioactive atoms of this element to be reduced by half. The half-life of the isotope beryllium- $11$ is $14$ seconds. A sample of beryllium- $11$ was first measured to have $800$ atoms. After $t$ seconds, there were only $50$ atoms of this isotope remaining. Write an equation in terms of $t$ that models the situation.
The strategy This problem involves the half-life of beryllium- $11$. So, to find the number of atoms of beryllium- $11$ remaining in the sample over time, we repeatedly multiply the original number of atoms, $800$, by $\dfrac{1}{2}$. Because of this, we know we can model the situation with an exponential expression of the form $ab^x$, where $a$ is $800$ and $b$ is $\dfrac12$. We now only need to find $x$, which represents the number of half-lives that have passed in $t$ seconds. Finding the exponent Suppose $t$ seconds have passed. Since the half-life of beryllium- $11$ is $14 $ seconds, the number of half-lives that passed is $\dfrac{t}{14}$. Writing an equation We can now replace $x$ in the original model with $\dfrac{t}{14}$. Therefore, the expression $800\cdot \bigg(\dfrac{1}{2}\bigg)^{\frac{t}{14}}$ models the number of atoms of beryllium- $11$ remaining in the substance after $t$ seconds. Since we know that $50$ atoms of this isotope remain after $t$ seconds, we can set the above expression equal to $50$. $50= 800\cdot \bigg(\dfrac{1}{2}\bigg)^{\frac{t}{14}}$ The answer An equation that models the problem is $800\cdot \bigg(\dfrac{1}{2}\bigg)^{\frac{t}{14}}=50$.